# Chapter 3 Stoichiometry

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CHAPTER 3

Topic Scopes:

Stoichiometry and

Solution Concentration

• Molarity, molality, parts per million &

percentage (w/w, w/v and v/v)

• Stoichiometry calculation

• Limiting reactant

• Theoretical yield, actual yield and

percentage yield

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Mole Concept

No. of Moles =

Molarity (M)

• Molarity (molar concentration) is the

number of moles of a solute that is

contained in 1 liter of solution

Mass (g)

molar mass (g/mol)

No. of Moles = Molarity (mol/L) volume (L)

Molarity (M) = Amount of solute (Mol)

Volume of solution (L)

• 1 mole contains 1 Avogadro’s number

(6.022 x 1023)

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Example: Saline Water

Concentration

Typical seawaters contain sodium chloride,

NaCl, as much as 2.7 g per 100 mL.

(Molar mass of Na = 22.99 g/mol; Cl = 35.45

g/mol; Mg = 24.30 g/mol)

(a) What is the molarity of NaCl in the saline

water?

(b) The MgCl2 content of the saline water is

0.054 M. Determine the weight (grams)

of MgCl2 in 50 mL of the saline water? 5

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Solution:

(a) Molar mass of NaCl = 22.99 +35.45

= 58.44 g mol-1

Moles of NaCl in 100 mL of saline water

= 2.7g /(58.44 g mol-1) = 0.046 mol

Molarity of saline water = Mol/L

= 0.046 mol /(100/1000)L = 0.46 M

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Chapter 4 — Intro—1

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Solution:

(b) Molar mass of MgCl2

= 24.30 + 2(35.45) = 95.20 g mol-1

Molality (m)

• Molality is the number of moles of solute

per kilogram (1000 g) of solvent

Moles of MgCl2 =

Molarity (M) x volume of solution (V)

Weight of MgCl2 in 50 mL of saline water

= (M x V) x MW

= 0.054 mol L-1 x (50/1000)L x 95.20 g mol-1

= 0.26 g

Molality (m) = Amount of solute (Mol)

Mass of solvent (kg)

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Percent Composition

(Concentration In Percentage)

Example:

A solution contains 118.5 g KI per liter of

solution. Calculate the concentration in (a)

% w/v & (b) % w/w. Given the density of the

solution at 25C is 1.078 g mL-1

(g)

(g)

(ml)

(ml)

Solution:

(a) % w/v = 118.5 g x 100%

1000 mL

= 11.85 % w/v

(g)

(ml)

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Units of Low Concentration

Solution:

(b) % w/w = 118.5 g x 1 mL x 100%

1000 mL 1.078 g

= 10.99 % w/w

• Parts per million, (ppm) is grams of solute

per million grams of total solution/ mixture

• ppm = mass of solute (g)

x 106

mass of sample(g)

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Unit of ppm: w/w = µg/g or mg/kg

w/v = µg/mL or mg/L

v/v = nL/mL or µL/L

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Chapter 4 — Intro—1

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Conversion of ppm to molarity

Units of Low Concentration

An aqueous solution contains 56 ppm SO2. Calculate

the molarity of the solution. (Molar mass of S = 32.06

g/mol; O = 16.00 g/mol)

• Parts per billion, (ppb) is grams of solute

per billion grams of total solution/ mixture

• ppb = mass of solute (g)

x 109

mass of sample(g)

Solution:

Molar mass of SO2= 32.06 + 2(16.00) = 64.06 g/mol

56 ppm 56 mg/L

Unit of ppb: w/w = ng/g or µg/kg

w/v = ng/mL or µg/L

v/v = nL/L

M

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56 mg

1g

1 mol

1L

1000 mg 64.06 g

8.74 10 4 M

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Reaction of Phosphorus with Cl2

Stoichiometry

Cl2

• The relationship

between the

quantities of

chemical reactants

and products

• Depend on the

principle of the

conservation of

matter

PCl3

P4

Notice the stoichiometric coefficients and the

physical states of the reactants and products

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Chemical Equations

• Depict the kind of reactants and products and

their relative amounts in a reaction

reactants

products

4 Al(s) + 3 O2(g) 2 Al2O3(s)

stoichiometric coefficients

• (s),(g),(l) – physical states of compounds

• (s) – solid, (g) – gas, (l) – liquid

(aq) – aqueous solution

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Law of The

Conservation of Matter

• States that matter can be

neither created nor

destroyed

• An equation must be

balanced

• It must have the same

number of atoms of the

same kind on both sides

of the equation

Lavoisier, 1788

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Chapter 4 — Intro—1

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Balanced Chemical Equation

Law of The

Conservation of Matter

12 Cl atoms...

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